Here $T$ is what, the natural Borel sigma-algebra? And the measures are probability measures? In my answer I assume these.

For this: $\mu(A)=\nu(f^{-1}(A))$, I might write $\mu = f(\nu)$ or maybe $\mu = f_*(\nu)$ and say that $\mu$ is the image of $\nu$ under $f$. Every probability measure on $2^\omega$ is an image of Lebesgue measure. Lebesgue measure is an image of a measure $\nu$ if and only if $\nu$ is atomless.

Here $T$ is what, the natural Borel sigma-algebra? And the measures are probability measures? In my answer I assume these.

For this: $\mu(A)=\nu(f^{-1}(A))$, I might write $\mu = f(\nu)$ or maybe $\mu = f_*(\nu)$ and say that $\mu$ is the image of $\nu$ under $f$. Every probability measure on $2^\omega$ is an image of Lebesgue measure. Lebesgue measure is an image of a measure $\nu$ if and only if $\nu$ is atomless.

[added Jun 14]

OK, we can change variables (except null sets) to get the following situation: $2^\omega$ is replaced by the square $[0,1] \times [0,1]$ and the map $f$ is the projection onto the first coordinate $[0,1]$. Now we want to know what are the measures on the square that project onto Lebesgue measure. Yes, indeed, there are lots of them.